Question

There is a class A having the virtual method print() and overloaded operator << defined as a friend function.

#include <iostream>
class A
{
public:
    double a1, a2;
    A(): a1(10.0), a2(10.0) {}

    virtual void print ( std::ostream * o = &std::cout ) const
    {
        *o << a1<< '\t' << a2 << '\n';
        }

    friend std::ostream & operator << ( std::ostream & o, const A &aa )
    {
        o << aa.a1 << '\t' << aa.a2 << '\n';
        return o;
    }
};

and analogously in derived class B

 class B : public A
 {
public:
    double b1, b2;
    B(): A(), b1(20.0), b2(20.0) {}

    virtual void print ( std::ostream * o = &std::cout ) const
    {
        A::print ( o );
        *o << b1<< '\t' << b2;
    }

    friend std::ostream & operator << ( std::ostream & o, const B &bb )
    {
        o << (A)(bb);
        o << bb.b1 << '\t' << bb.b2 << '\n';
        return o;
    }
 };

I have the following questions:

1] Is there any way how to pass a pointer to ostream object with default parameter so as operator << correctly replaces the print() method? This overloading is wrong

friend std::ostream & operator << ( std::ostream * o= &std::cout, const A &aa )

2] I am not sure, if this line calling operator of the parent class A in derived class B is correct?

 o << (A)(bb);

3] Is there any better way how to overload operator << without “friend” declaration?

Thanks for your help….

Answer 1

You can do it like this without friendship:

#include <iostream>
class A
{
    double a1, a2;

public:
    A(): a1(10.0), a2(10.0) {}

    virtual void print ( std::ostream * o = &std::cout ) const
    {
        *o << a1 << '\t' << a2;
    }
};

std::ostream & operator << ( std::ostream & o, const A &aa )
{
    o << "( )";
    aa.print(&o);
    return o << " )";
}

and then class B doesn’t a separate version of operator<<, this one will be found, and call B::print when you pass it a B instance.